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Chemistry tutorial

  1. Feb 5, 2004 #1
    Is there a webpage with illustrations and well formed explanations on such as mol, g/mol, Mm etc etc? The Norwegian word for this part of chemistry is støkiometri. I believe it is something like stochiometry in english even though that sounds too odd.

    I had a test on this today and I believe I am near failing. I do know much about this subject, but everything dropped out of my head when I got the test.

    I am grateful for any help!
  2. jcsd
  3. Feb 5, 2004 #2
    from google (on the proper spelling of the word 'stochiomistry' as it is there) http://www.chem.tamu.edu/class/majors/tutorialnotefiles/stochiometry.htm [Broken]

    EDIT (Pssst...I didn't read it...)
    Last edited by a moderator: May 1, 2017
  4. Feb 5, 2004 #3


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    Don't you have a chemistry book?

    Try searching for Ask Antoine on google. It probably has tutorials on this subject.
  5. Feb 5, 2004 #4
    Ironic that that site's url spells stoichiometry wrong yet in the site itself it spells it correctly.

    The thing with stoichiometry is that it is very much like balancing equations, but instead of the balancing process itself, it is dealing with the amount of substances in the equation.

    Mainly it is converting... "how many moles of X will it take to react with moles of Y". I will provide a very basic example that encompasses a broad variety of cases.

    Suppose you had Tin (Sn) which reacts with Hydrogen Fluoride (HF) that yields Tin Fluoride (SnF2) and Hydrogen Gas. You wanted to know how many grams of SnF2 are produced from 30.00 g of HF.

    The equation goes like this: Sn + HF -> SnF2 + H2
    Now, to work with it, it must be balanced: Sn + 2HF -> SnF2 + H2

    Let's start with what is given. 30.00 g of HF needs to be converted to moles (what I call the mole superhighway, can go anywhere). From there it multiplies with mole ratio of SnF2 (what your trying to find) divided by HF (HF is on bottom to cancel it out). Then the moles of SnF2 is converted back into grams and the problem is solved.

    On paper, it should look like this:
    30.00 g HF x 1 mol HF x 1 mol SnF2 x 156.71 g SnF2 = answer
    .....................s1------ --s2------ --s3---------
    .....................20.01 g HF 2 mol HF 1 mol SnF2

    (dont mind the periods, those are just space fillers to make the thing look right)

    step 1 - 20.01 g HF is molar mass of HF (mass of H + mass of F)
    step 2 - This is the conversion part, and the numbers on numerator and denominator correspond to the balanced equation above (Sn + 2HF -> SnF2 + H2). Note that these numbers only come into play when converting between two different compounds.
    step 3 - Converting back into grams by multiplying it by its molar mass (in this case 156.71).

    The correct answer should be 117.5 g SnF2. Hope this helps.
    Last edited: Feb 5, 2004
  6. Feb 8, 2004 #5


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    Understanding stoichiometry is a little different than understanding the basic definitions of what a mole or molar concentration are. You need to understand the basic definitions before you can understand all the rest.

    To start off, a mole (abbreviated mol...never understood why we needed an abbreviation of such a small word), is just a counting number. It would be just like saying something is a million or a billion of something, except it's a strange unit because it doesn't refer to 1 followed by some number of zeros, it refers to 6.022 x 10^23, and it is used for counting numbers of molecules.

    Then, molarity refers to a concentration. The units are the number of moles in a liter of solution. So, if a solution of something is 1 molar (or 1 M), that means in 1 liter, there are 6.022 X 10^23 molecules of that something.

    The molecular weight of something is expressed as grams/mole (g/mol). That means if you were to sit down and count out 6.022 x 10^23 molecules (1 mole) of something onto a gram balance, the molecular weight is the gram weight you'd read on the balance. Molecular weight isn't really determined this way, but that's what it ultimately means.

    Once you understand these three basic concepts, balancing stoichiometric equations is just algebra.
  7. Feb 8, 2004 #6
    I appreciate your efforts, but I allready know all this. I delivered the tesst papers with the problems and I will post them when I get the test back. I had no idea on what to do and what to calculate and which formulas to use. I want to figure this out!

    I will post the tasks soon!
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